Math, asked by 123bakisabfree, 1 year ago

help me to understand the steps given in the box

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Answers

Answered by 2028hacker
0
As per the data given in the root the square of 10 is opened first
and 4,2,600 are multiplied.
In the next step 100( square of 10) is added to 4800(product of multiplicants) resulting in 4900 which is the square of 70.
Hence it has been solved

Can I be the brainliest please
Answered by rakeshmohata
6
Hope u like my process
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It's better to take 2 common from the terms and write it in a form of

=> x² + ax + c = 0
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So..

 =  > 2 { {y} }^{2}  + 10y - 600 = 0 \\  \\  =  >  \blue{ {y}^{2} + 5y - 300  } =  \blue0 \\  \\  =  >   \blue{{y}^{2} + 20y - 15y - 300} =  \blue0 \\  \\  =  > \blue{ y(y + 20) - 15(y + 20)} =  \blue0 \\  \\  =  >  \blue{(y - 15)(y + 20)} =  \blue0 \\  \\  \underline{thus} \\  \\  =  >  \boxed{y =  \green{15}} \\  \\ or \\  \\  =  >  \boxed{y =   \green{- 20}}

And.. If using the form of

 =  > x =  \blue{ \frac{ - b +  \sqrt{ {b}^{2}  + 4ac} }{2a}} ...or...  \blue{\frac{ - b -  \sqrt{ {b}^{2} - 4ac} }{2a} }
of equation of type ax² + bx + c =0

So..

here \:  \:  \:  \bf \: a \: = 1 \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \bf\: b  = 5 \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bf c =  - 300 \\

Thus,..
 =  > y = \green{  \frac{ - 5 +  \sqrt{ {5}^{2} - 4 \times 1 \times ( - 300) } }{2 \times 1} } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:   =  \green{ \frac{ - 5 +   \sqrt{25 + 1200}  }{2} } =  \green{ \frac{ - 5 \sqrt{1225} }{2} } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:   =  \green{ \frac{ - 5 + 35}{2} } =  \green{ \frac{30}{2} } =  \underline{ \green{15}} \\  \\ and. \\  \\   = > y =   \green{\frac{ - 5 -  \sqrt{ {5}^{2} - 4(1)( - 300) } }{2 \times 1}}  \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = \green{  \frac{ - 5 - 35}{2} } =  \green{ -  \frac{40}{2} } =  \underline{ \green{ - 20}}

Thus

The required values of y are..

 \boxed{ =  > y =  \bold{ \underline{ \orange{15} \:  \: or \:  \:  \orange{ - 20}}}}
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Hope this is ur required answer

Proud to help you
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